ĐK để pt có nghiệm là x>0.\(\Rightarrow\sqrt{x^2-x-1}
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GPT \(2x+3+\sqrt{4x^2+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}\)
GPT : \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
gpt: \(\sqrt{9x^2+16}=2\sqrt{2x+4}+4\sqrt{2-x}\)
gpt: \(\sqrt{9x^2+16}=2\sqrt{2x+4}+4\sqrt{2-x}\)
a) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
b) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
c) \(\sqrt{x^2+6x-9}-2\sqrt{x^2-2x+1}+\sqrt{x^2}=0\)
gpt: \(\sqrt{9x^2+16}=2\sqrt{2x+4}+4\sqrt{2-x}\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
L2
26 tháng 10 2021 lúc 9:41
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
GPT
a) \(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\)
b) \(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\sqrt{2}\)